2015 AMC 10A Problem 13

Attempt Problem 13 of the 2015 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 10A solutions, or check the answer key.

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13.

Claudia has 1212 coins, each of which is a 55-cent coin or a 1010-cent coin. There are exactly 1717 different values that can be obtained as combinations of one or more of her coins. How many 1010-cent coins does Claudia have?

33

44

55

66

77

Answer: C
Concepts:moneycounting integers in a range
Difficulty rating: 1480
Solution:

Let the number of 55-cent coins be xx and the number of 1010-cent coins be 12x.12 - x.

If x=0,x=0, the only possible values are 10,20,,12010,20,\ldots,120 cents, giving just 1212 values. Hence x>0.x>0. Having at least one 55-cent coin then makes every multiple of 55 from 55 through the total value 5x+10(12x)=1205x5x+10(12-x)=120-5x obtainable.

There are 24x24 - x such multiples of 5,5, which means that x=7x = 7 to get 1717 possible different values.

The number of 1010-cent coins is therefore 127=5.12 - 7 = 5.

Thus, C is the correct answer.

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